How to invert a full matrix?

Question

I have a matrix

            X1         X2        X3        X4        X5        X6           G1            G2
X1            5         0         0         0         0         0            5            0 
X2            0         5         0         0         0         0            5            0
X3            0         0         5         0         0         0            5            0
X4            0         0         0         5         0         0            0            5
X5            0         0         0         0         5         0            0            5  
X6            0         0         0         0         0         5            0            5  
G1            5         5         5         0         0         0           15            0  
G2            0         0         0         5         5         5            0           15  

There are dependence lineal in the two last rows and columns, I try to delete one row and one column (for example G1), but the resulting matrix still has a linear dependence, so I can't invert the matrix.

Can anyone help me please?

Answer 1

The symmetric matrix m, shown reproducibly in the Note at the end, is singular which is implied by its having at least one zero eigenvalue (in fact, it has two):

eigen(m)$values
## [1]  2.000000e+01  2.000000e+01  5.000000e+00  5.000000e+00  5.000000e+00
## [6]  5.000000e+00 -4.671248e-24 -4.671248e-24

Thus, it cannot be inverted; however, you can take the Moore Penrose generalized inverse using ginv from the MASS package. A generalized inverse does satisfy the relation shown in the last line of code below. Note that the MASS package comes with R so you don't need to install it.

library(MASS)

gm <- ginv(m)
all.equal(m %*% gm %*% m, m)
## [1] TRUE

Note that there are two linear dependencies owing to the two zero eigenvalues. By inspection we note that the sum of the first 3 rows of m equals the 7th row and the sum of the 4th through 6th rows equals the 8th row so dropping rows 7 and 8 and the corresponding columns due to symmetry we see that the upper left 6x6 submatrix is invertible:

solve(m[-(7:8), -(7:8)])

Alternately we can find constraints by inspection of the eigenvectors of m.

Note

Lines <- "X1         X2        X3        X4        X5        X6           G1            G2
X1            5         0         0         0         0         0            5            0 
X2            0         5         0         0         0         0            5            0
X3            0         0         5         0         0         0            5            0
X4            0         0         0         5         0         0            0            5
X5            0         0         0         0         5         0            0            5  
X6            0         0         0         0         0         5            0            5  
G1            5         5         5         0         0         0           15            0  
G2            0         0         0         5         5         5            0           15"
m <- as.matrix(read.table(text = Lines))